Theorem nss 4028

Description: Negation of subclass relationship. Exercise 13 of [TakeutiZaring] p. 18. (Contributed by NM, 25-Feb-1996.) (Proof shortened by Andrew Salmon, 21-Jun-2011.)

Assertion

Ref	Expression
nss	⊢ (¬ 𝐴 ⊆ 𝐵 ↔ ∃𝑥(𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵))

Distinct variable groups:   𝑥,𝐴   𝑥,𝐵

Proof of Theorem nss

Step	Hyp	Ref	Expression
1		exanali 1858	. . 3 ⊢ (∃𝑥(𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵) ↔ ¬ ∀𝑥(𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵))
2		df-ss 3948	. . 3 ⊢ (𝐴 ⊆ 𝐵 ↔ ∀𝑥(𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵))
3	1, 2	xchbinxr 335	. 2 ⊢ (∃𝑥(𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵) ↔ ¬ 𝐴 ⊆ 𝐵)
4	3	bicomi 224	1 ⊢ (¬ 𝐴 ⊆ 𝐵 ↔ ∃𝑥(𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵))

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 206   ∧ wa 395  ∀wal 1537  ∃wex 1778   ∈ wcel 2107   ⊆ wss 3931

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1794  ax-4 1808

This theorem depends on definitions:  df-bi 207  df-an 396  df-ex 1779  df-ss 3948

This theorem is referenced by:  grur1  10842  psslinpr  11053  reclem2pr  11070  mreexexlem2d  17659  prmcyg  19880  filconn  23837  alexsubALTlem4  24004  wilthlem2  27048  shne0i  31395  erdszelem10  35164  fundmpss  35726  ntrneineine1lem  44059  nssrex  45048  nssd  45067  nsssmfmbf  46751